Let us start with two facts and a remark.
Fact 1. Let $(X,\mathcal{T})$ be a topological space and let $f \colon (X,\mathcal{T}) \to
\left[{-}\infty,{+}\infty\right]$. Then $f$ is lower semicontinuous if and only if, for every $\xi \in \mathbb{R}$, the lower level set $f^{-1}(\left[{-}\infty,\xi\right])$ is closed. Here, by lower semicontinuity, I mean: for every $x \in X$ and for every $\xi \in \left]-\infty,f(x)\right[$, there exists a neighborhood $V$ of $x$ in such that $(\forall y \in V)\; f(y) > \xi$.
Remark Lower semicontinuity goes with the topology on the domain of $f$. In particular, in your question, lower semicontinuous means "$f$ is lower semicontunuous wrt to the strong topology" whereas "weakly lower semicontinuous" means "$f$ is lower semicontinuous wrt to the weak topology on $H$." So I guess there is no way to avoid weak topology in the proof as it directly relates to the topologies on the domain.
Fact 2. Let $C$ be a convex subset of $H$ (in your question). Then $C$ is closed in the topology induced by the hilbertian norm of $H$ if and only if $C$ is closed in the weak topology.
Returning to your question and assume that $f$ is lower semicontinuous w.r.t the strong topology (induced by the norm of $H$) and that $f$ is convex. We must show that $f$ is weakly lower semicontinuous, i.e., $f$ is continuous when $H$ is equipped with the weak topology. Let us use Fact 1 to do this, i.e., take $\xi \in \mathbb{R}$ and show that $f^{-1}(\left[{-}\infty,\xi\right])$ is weakly closed. Since $f$ is convex, the set $f^{-1}(\left[{-}\infty,\xi\right])$ is convex. On the other hand, since $f$ is lsc w.r.t to the strong topology, the set $f^{-1}(\left[{-}\infty,\xi\right])$ is closed in the strong topology by Fact 1. Altogether, Fact 2 implies that it is indeed weakly closed.
So, we have shown that, for every $\xi \in \mathbb{R}$, the set $f^{-1}(\left[{-}\infty,\xi\right])$ is closed in the weak topology. In view of Fact 1, we conclude that $f$ is weakly lsc, i.e., lower semicontinuous when $H$ is equipped with the weak topology.
If $f$ is a continuous linear functional then $f(x_n) \to f(x)$. Let $0<\epsilon <\Re f(y)-\Re f(x)$. There exists $N$ such that $|f(x_k)-f(x)| <\epsilon$ for all $ k \geq n$. This implies that $|f(z)-f(x)| \leq \epsilon$ for all $z \in K_N$. Hence, $\Re f(z)<\epsilon +\Re f(x) <\Re f(y)$ if $z \in K_N$.
We get the desired contradcition by taking $z=y$.
Best Answer
Since $x_n$ converges weakly to $x$, one has for every $a \in H$. $$\lim_{n \to \infty} \langle a, x_n \rangle = \langle a, x \rangle.$$ Consequently, one has $$\lim_{n \to \infty} \sup_{z \in K_n} |\langle z, a\rangle - \langle x, a \rangle| = 0.$$ This implies for $y \in \bigcap K_n$, one has for every $a \in H$ $$\langle y, a \rangle = \langle x, a \rangle,$$ which yields $y = x$.